EXPONENTIATIONS OVER THE UNIVERSAL ENVELOPING ALGEBRA OF sl2(C)
SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2
Abstract. We construct, by model-theoretic methods, several exponentiations on the universal enveloping algebra U of the Lie algebra sl2(C).
MSC: 16S30, 17B10, 03C60. Key words: universal enveloping algebra, exponential map, asymptotic cone
1. Introduction
Consider the Lie algebra M2(C) and the Lie (sub)algebra sl2(C) of all 2 × 2 trace zero matrices with complex entries. Recall that a standard basis of sl2(C) (as C-vectorspace) is 0 1 0 0 1 0 given by: x = , y = , h = = diag(1, −1). The generators 0 0 1 0 0 −1 x, y, h satisfy the relations: [h, x] = 2x,[h, y] = −2y,[x, y] = h, where [u, v] is the usual commutator of u and v . For each positive integer λ, we consider the finite-dimensional simple sl2(C)-module Vλ of dimension λ + 1 and the (matrix) Lie algebra Mλ+1(C) (the endomorphism ring of Vλ, viewed as a C-vectorspace) and take the exponential maps from Mλ+1(C) into the linear group GLλ+1(C). (In section 3, we recall some properties of these exponential maps.) We connect these exponentials to the universal enveloping algebra U of sl2(C) (whose definition and algebraic properties are described in Section 4). We will use some basic facts on the representation theory of this associative algebra (and its analogue over any algebraically closed field of characteristic 0). It has been studied from a model theoretical point of view by [11] and then by [12, 16, 17]. Using on one hand the concrete exponential maps defined on the matrix rings Mλ+1(C) and on the other hand the universal property of U, we define a sequence of exponential maps indexed by λ from U to GLλ+1(C). We describe some of the properties of these maps, which we have formalized (in section 2) by defining the notion of a non-commutative exponential ring (generalizing the commutative case) and we explicitly calculate elements lying in their kernels (respectively images). Then, we show that U embeds into any non- Q principal ultraproduct Mλ+1(C) and we define an exponential map EXP from U to V Q any non-principal ultraproduct V GLλ+1(C) of the groups GLλ+1(C), where V is a non principal ultrafilter on ω. We show that (U, EXP ) is a non-commutative exponential ring, and we explicitly calculate a part of the kernel of EXP . Note that a formal exponential map exp was previously defined in the completion Uˆ of U ([22]), on the ideal on Uˆ generated by the generators of U; in section 7, we will compare the two approaches.
Date: May 5, 2010.
1 The first author was partially supported by MODNET Research Training Network in Model Theory 2 Research Director at the ”Fonds de la Recherche Scientifique-F.N.R.S.”. 1 2 SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2 Q We go on to endow U with a topology using a norm in V Mλ+1(C) which takes its values in a non-standard ultrapower of R, and we show that the exponential map EXP is continuous and that the subgroup generated by EXP (U) is a topological group. Finally, by considering another norm on each Mλ+1(C), and the asymptotic cone relative to these norms and a non-principal ultrafilter V on ω, we embed U in a complete metric space and show that U has a faithful continuous action on that space.
2. Preliminaries on formalism Let us set up the languages we need. Let Lg := {·, 1} be the language of groups. Let L := {+, −, ·, 0, 1} be the language of (associative) rings, and let Ll := {+, −, [·, ·], 0} be the language of Lie rings. For a ring R, let Lm,R := {+, −, 0, ·r; r ∈ R} be the language of right R-modules. For the language of R-algebras, where R is a commutative ring, we will choose the expansion LAlg of L, a two-sorted language with a sort for a ring R, a sort for an algebra A (associative or not) and a scalar multiplication map from A × R to A, where A is either a L-structure or a Ll-structure and R is a L-structure. For the language of Lie K-algebras, where K is a field, we will choose either LLie := Ll ∪ Lm,K or the two-sorted language LAlg. Note that for the former we omit reference to K when it is understood. We will assume that K is a field of characteristic 0 and is complete with respect to a nontrivial absolute value.
Let TK be the theory of K-vector spaces in Lm,K . Let TL be the theory of Lie K-algebras in the language LLie, namely
(1) TK , (2) [·, ·] is a K-bilinear map from L × L to L, (3) ∀x1 [x1, x1] = 0, (4) ∀x1, ∀x2 ∀x3 ( [[x1, x2], x3] + [[x2, x3], x1] + [[x3, x1], x2] = 0 ).
3. Axioms for semisimple Lie algebras We will translate, into model-theoretic terms, the basic results on existence and unique- ness of a semi-simple Lie K-algebra with a given reduced abstract root system Φ ([13], chapter 18.2).This is not essential for the present paper, but may be of interest in future generalizations. Recall that Φ is a subset of an Euclidean space E endowed with a positive definite (β,α) symmetric bilinear form (., .). Denote by < β, α >:= 2. (α,α) . For a root system Φ these values are integers. For x ∈ L, let ad x be the linear transformation of L sending y ∈ L to [x, y].
Proposition 3.1. The theory of any semi-simple Lie algebra L with given reduced root system Φ( and inner product on it) is axiomatisable in LAlg by the set TΦ of axioms below. Moreover each TΦ is ℵ1-categorical.
(1) TAlg the theory of K-algebras in LAlg over some field K; (2) The scheme of axioms expressing that K is an algebraically closed field of charac- teristic 0; EXPONENTIATIONS OVER THE UNIVERSAL ENVELOPING ALGEBRA OF sl2(C) 3
(3) (the αj are the elements of the root system) ∃h1 · · · ∃h` ∃e1 · · · ∃e`∃e−1 · · · ∃e−` V [ 1≤i,j≤`[hi, hj] = 0 V V 1≤i≤`[ei, e−i] = hi & 1≤i6=j≤`[ei, ej] = 0 V V 1≤i,j≤`[hi, ej] =< αj, αi > ·ej & 1≤i,j≤`[hi, e−j] = − < αj, αi > ·e−j V −<αj ,αi>+1 1≤i6=j≤`(ad ei) (ej) = 0 V −<αj ,αi>+1 1≤i6=j≤`(ad e−i) (e−j) = 0 P P P & ∀x ∃k1 ∈ K · · · ∃k3` ∈ K x = 1≤i≤` ki.hi + 1≤i≤` k`+j.ej + 1≤j≤` k2`+j.e−j ]. Proof: Serre’s work tells us that given a root system Φ and a field of characteristic 0, there exists a unique Lie algebra L that can be presented by these relations and that it is semi- simple. The second statement follows from the fact that if L is a model of these axioms of cardinality ℵ1, then it is a Lie algebra over an algebraically closed field F of characteristic 0 of cardinality ℵ1. 2 Question 3.1. Is the theory of any semi-simple Lie C-algebra L with given root system Φ finitely axiomatisable in LLie modulo TC? Let Axiom (3’) be got from Axiom (3) by deleting the last part where we quantify over x. Let L be a model of (1),(2) and (3’), and let L0 be the Lie subalgebra generated by the elements hi, ej, e−j satisfying the above relations. Then, Serre’s theorem tells us that L0 is a semi-simple finite-dimensional Lie algebra with root system Φ and Cartan subalgebra generated by hi, 1 ≤ i ≤ `. Then can we add an axiom that forces L to be equal to L0?
We could have also worked in the language Ll since any semi-simple Lie algebra has a basis with integral structure constants (a Chevalley basis). We do not pursue this matter here.
4. Exponential rings
Let LE := L ∪ {E} (respectively LAlg,E := LAlg ∪ {E}) where E is a unary function symbol. We will introduce the notion of (non commutative) exponential ring generalizing the commutative case (see for instance [7]).
Definition 4.1. Let (R,E,G) be a two-sorted structure with R an L-structure, G a Lg- structure and E a map from R to G. We will say that (R,E,G) is an exponential ring if R is an associative ring with 1, G a (multiplicative) group and if E : R → G satisfies the following axioms: (1) E(0) = 1, (2) ∀x E(x).E(−x) = 1, (3) ∀x ∀y (x.y = y.x → E(x + y) = E(x).E(y)). If in addition R is a K-algebra, then (R,K,E,G) is an exponential K-algebra if (R,K) is a LAlg-structure such that the reduct (R,E,G) is an exponential ring, the LAlg-reduct (R,K) a K-algebra and
∀k1, k2 ∈ K ∀x ∈ RE(k1.x).E(k2.x) = E((k1 + k2).x). Note that this last axiom together with (1) implies (2) above. One recovers the classical case by taking G the group of units of R, by assuming that R is a commutative ring and then we revert to the one-sorted LE-structure (R,E). In the case we deal with an exponential K-algebra, we will get that (K,E) is an exponential field. 4 SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2
5. A natural exponential map over Mλ+1(C) Consider the field C of complex numbers and, for a fixed natural number λ, the associative C-algebra Mλ+1(C) of all (λ + 1) × (λ + 1) matrices with coefficients in C (with the matrix multiplication · as the underlying operation). It is also a Lie C-algebra with the bracket ∗ [A, B] := A · B − B · A (see [8, 13]). For A ∈ Mλ+1(C), denote by A the conjugate of the transpose of A, by tr(A) the trace of A , and finally by det(A) its determinant. We will denote by Diagλ+1(C) (respectively UTλ+1(C)) the subset of all diagonal matrices (respectively upper triangular matrices) in Mλ+1(C). Recall that on the Lie algebra Mλ+1(C), we have a Hermitian sesquilinear form (·, ·)λ+1 ∗ P ¯ defined by (A, B)λ+1 := tr(B · A) = i,j Aij · Bij, where A, B ∈ Mλ+1(C), its values are in C ([21] page 9). The Frobenius norm (denoted by F -norm) associated with it, is defined 2 as follows: kAkF,λ+1 := (A, A)λ+1. We use this norm systematically later. In addition to the triangle inequality and submultiplicativity (from which multiplication is continuous for the norm topology) the F - norm satisfies the Cauchy-Schwartz inequality ([21] page 10). Note that for diagonizable matrices, the F - norm is the square root of the sum of the squares of the norms of the eigenvalues of the matrix. λ+1 There are many norms on C , all giving the same topology. For example, on C we λ+1 have the usual norm | · |, inducing on C the norm (”the 2-norm”) whose value is the square root of the sum of the squares of the absolute values of the entries. This norm, and the Frobenius norm , are both instances of Schatten 2-norms. When we refer later to norms, it will be to such norms, unless we deal explicitly with operator norms. λ+1 λ+1 We consider the elements of Mλ+1(C) as linear operators φ from (C , k·k1) to (C , k· λ+1 k2). Then, for any ordered pair of norms on C there is a corresponding operator norm on Mλ+1(C). Later, when we consider ultraproducts of the Mλ+1(C) we will return to discussion of such norms. We will use operator norms only with reference to Schatten 2-norms. From now on, we will assume that Mλ+1(C) is equipped with a fixed norm k·k satisfying the Cauchy-Schwartz inequality. The topology on Mλ+1(C) is independent of the norm, but in discussing convergence of series we will appeal to the fixed norm. If A is any matrix in Mλ+1(C), one defines ([21] 1.1) the matrix exponential of A, denoted by exp(A), as the power series: ∞ X An A2 A3 (1) exp(A) = = I + A + + + ... n! λ+1 2 3! n=0 where Iλ+1 denotes the (λ + 1) × (λ + 1) identity matrix. This exponential series converges in norm for all matrices, so the exponential of A is well-defined. If A is a 1 × 1 matrix, that a a is, a scalar a of the field C), then exp(A) = e where e denotes the ordinary exponential of the element a ∈ C. Recall that the matrix exponential satisfies the following properties:
Proposition 5.1. Let A, B ∈ Mλ+1(C) and a, b ∈ C we have: (i) exp (0λ) = Iλ+1, where 0λ+1 denotes the zero matrix in Mλ+1(C); (ii) exp (aA) · exp (bA) = exp ((a + b)A); (iii) exp (A) · exp (−A) = Iλ+1; (iv) for A and B commuting, exp (A + B) = exp (A) · exp(B); (v) for an invertible matrix B, exp (BAB−1) = Bexp (A)B−1; (vi) det(exp (A)) = exp (tr(A)). EXPONENTIATIONS OVER THE UNIVERSAL ENVELOPING ALGEBRA OF sl2(C) 5
(vii) If no two eigenvalues of A have a difference belonging to 2π.i.Z, then there exists a neighbourhood of A on which exp is injective. The exponential is injective on the ball of radius log(2) around the origin, for the Frobenius norm. Proof. See [21] Proposition 1 (b), (c), (d), Proposition 3 in section 1.1 and Proposition 7 in section 1.2. See also [1, Chapter 3]. It remains to prove (vi). By (v), in order to prove (vi)one may assume without loss of generality that A is in Jordan normal form. So A can be written as a sum of a diagonal matrix D and a nilpotent matrix N with N and D commuting. So, by (iv), exp(D+N) = exp(D)·exp(N). Therefore, det(exp(A)) = det(exp(D)) · det(exp(N))=det(exp(D)). 2 For non commuting matrices A and B, the equality exp(A + B) = exp(A) · exp(B) need not hold. In that case, the Baker-Campbell-Hausdorff formula can be used to express exp(A) · exp(B) (see [21, Section 1.3]). Now, using the matrix exponential, one defines the exponential map
exp : Mλ+1(C) → GLλ+1(C): A 7→ exp(A), and rephrasing, (part of) Proposition 5.1, we get that (Mλ+1(C), exp, GLλ+1(C)) is an exponential C-algebra. Moreover, the map exp is surjective from Mλ+1(C) to GLλ+1(C). (Every invertible matrix can be written as the exponential of some other matrix ([21] page 21).) For future use, we recall some methods for explicitly calculating matrix exponentials.
Diagonalizable case. If a matrix A ∈ Mλ+1(C) is diagonal A = diag(a1, a2, . . . aλ+1), then its exponential can be obtained by just exponentiating every entry on the diago- nal: exp(A) = diag(ea1 , ea2 , . . . , eaλ+1 ). This also allows one to exponentiate any diagonalizable (so-called semisimple) matrix −1 S ∈ Mλ+1(C) . If S = BDB where B is invertible and D is diagonal, then, accord- ing to the property (v) in Proposition 5.1, we have that exp(S) = Bexp(D)B−1 and the exponential of the matrix D is calculated as above. q Nilpotent case. Recall that a matrix N ∈ Mλ+1(C) is nilpotent if N = 0 for some positive integer q ( without loss of generality ≤ λ + 1). In this case, the matrix exponential exp(N) can be computed directly from the series expansion (expressed by (1)), as the series terminates after a finite number of terms: N 2 N q−1 exp(N) = Iλ+1 + N + 2 + ... + (q−1)! .
General Case. Since any matrix A ∈ Mλ+1(C) can be expressed uniquely as a sum A = S + N where S is diagonalizable, N is nilpotent and S · N = N · S, then the ex- ponential of A can be computed by using the property (iv) of Proposition 5.1 and by reducing to the previous two cases, so: exp(A) = exp(S + N) = exp(S) · exp(N) . Note that this uniqueness easily translates, via quantifier elimination for algebraically closed fields, into a constructible version in the sense of algebraic geometry. We will need a more thorough description of Ker(exp). It is easy to see that the map exp is not injective. For instance, consider a nonzero diagonal matrix Iλ+1 6= D ∈ Mλ+1(C), d1 d2 d D = diag(d1, d2, . . . dλ+1), with its matrix exponential, exp(D) = diag(e , e , . . . , e λ+1 ). Then, exp(D) ∈ Ker(exp) if and only if the entries of D belong to the kernel of the standard complex exponential map , so if and only if d1, d2, . . . dλ+1 ∈ 2πiZ. 6 SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2
Lemma 5.2. If the matrix A ∈ Mλ+1(C) belongs to the kernel of exp, then A is diagonal- isable and its eigenvalues lie in the kernel of the exponential function e in C. Proof: In order to determine whether A ∈ Ker(exp), by Proposition 5.1 (v), since exp(B−1 · −1 −1 A · B) = B · exp(A) · B, we have that expA = Iλ+1 if and only if exp(B · A · B) = Iλ+1, for any invertible matrix B ∈ Mλ+1(C). Since C is algebraically closed, A is conjugated to a matrix in Jordan normal form, which can be written as a sum of a diagonal matrix D and a nilpotent matrix N with N and D commuting. So, by Proposition 5.1 (iv), exp(D + N) = exp(D) · exp(N).Now exp(N) is unipotent. So if exp(D + N) = Iλ+1, then exp(D) = exp(−N) , so the diagonal matrix exp(D) = Iλ+1 . Thus the eigenvalues of D are periods of the exponential on the complex numbers. Also, exp(N) must be Iλ+1. Finally, a simple calculation with the polynomial exp(N) (in N) gives N = 0λ+1. We conclude that the kernel of exp consists of the diagonalizable matrices with complex periods as eigenvalues.
Proposition 5.3. Each associative Lie algebra (Mλ+1(C), exp) viewed as an LE- structure x is bi-interpretable with the exponential field (C, e ). (λ+1)2 Proof: (See [18]). We embed Mλ+1(C) in the direct product C . Then since C is algebraically closed, any matrix is conjugate to a matrix in Jordan normal form, namely D + N where D and N commute, D is a diagonal matrix and N a nilpotent matrix with N λ+1 = 0. By Proposition 5.1, exp(D + N) = exp(D) · exp(N). Furthermore, exp(N) = 1 + N + ··· + N λ. For the other direction of interpretability, see [18]. 2
However, note that the class of rings {Mλ+1(C); λ ∈ ω} is undecidable since the class of their invertible elements {GLλ+1(C); λ ∈ ω} is undecidable (one interprets uniformly in λ a class of finite models whose theory is undecidable) and this implies that the theory of any non-principal ultraproduct of the Mλ+1(C) is undecidable. Moreover, note that one may replace C by an arbitrary field, and the group GLλ+1(C) by other algebraic groups like SLλ+1(C) (see [9]).
6. The universal enveloping algebra of sl2(C)
Recall that the universal enveloping algebra U of sl2(C) is an associative C-algebra (hence, equipped by a Lie algebra structure) together with a canonical mapping σ which is a Lie algebra homomorphism σ : sl2(C) → U such that, if R is any associative C-algebra and f : sl2(C) →R is a Lie algebra homomorphism, then there exists a unique algebra homomorphism Θ : U → R sending 1 to 1 and such f = Θ ◦ σ (see [6] chapter 2, sections 1, 2).
Diagram 6.1. Let us choose as R the Lie algebra M2(C) and as f the Lie algebra homo- morphism f1 : sl2(C) → M2(C), so there exists a unique algebra homomorphism Θ1 : U → M2(C) such that (according to what just said above) the following diagram commutes.
σ - sl2(C) U
f 1 Θ 1 ? © M2(C) EXPONENTIATIONS OVER THE UNIVERSAL ENVELOPING ALGEBRA OF sl2(C) 7
Since the canonical mapping σ of sl2(C) into U is injective ([6] Proposition 2.1.9), from now on we will identify every element of sl2(C) to its canonical image in U. By using this universal property of U, we can construct an exponential map over U. Let us define the exponential map from U to GL2(C) as follows:
Θ1 - exp- EXP1 : U M2(C) GL2(C)
EXP1(α) = exp (Θ1(α)) ∀α ∈ U.
So, the values of EXP1(U) are in GL2(C) and the restriction of EXP1 to sl2(C) coincides with the exponential map exp : sl2(C)→GL2(C) (viewing sl2(C) ⊂ M2(C)), previously defined (see (1)). Note that the image of the restriction of exp to sl2(C) is included in SL2(C) (see Proposition 5.1 (vi)). Clearly (U, EXP1, GL2(C)) is an exponential algebra. Let c = 2x · y + 2y · x + h2 be the Casimir element of U, where x, y, h are the generators of sl2(C). c generates the center of U. Let us calculate EXP1(c). First, let 0 1 0 0 0 0 0 1 Θ (c) = Θ (2x · y + 2y · x + h2) = 2 · + 2 · + 1 1 0 0 1 0 1 0 0 0 1 0 0 0 + (diag(1, −1))2 = 2 + 2 + diag(1, 1) = diag (3, 3) . 0 0 0 1
3 3 By using the universal property of U, we have that EXP1(c) = exp (Θ1(c)) = diag(e , e ).
Now we want to describe the map EXP1 on U. Recall that U is a Z-graded algebra with grading gr(x) = 1, gr(y) := −1 and so gr(h) = 0; a m-homogeneous element u ∈ U is an element such that gr(u) = m, m ∈ Z. So U decomposes as a direct sum of m-homogeneous components Um consisting of m-homogeneous elements, m ∈ Z,
U = ⊕m∈ZUm. Furthermore, every m-homogeneous component satisfies the following relation, depending on whether m is positive or negative: m m Um = x U0 = U0x for every positive integer number m |m| |m| Um = y U0 = U0y for every negative integer number m where, as well-described in [11], the 0-homogeneous component U0 coincides with the ring of polynomials C[h, c] with coefficients in C and variables the diagonal matrix h and the Casimir element c. Let um be an element in Um for m a positive integer. So, um = m m 2 m m 2m uo · x = x · v0, for some u0, v0 ∈ U0, and um = (u0 · x ) · (x · v0) = u0 · x · v0. 2 2m 2.m Applying Θ1 to um, we can see that Θ1(um) = Θ1(u0 · x · v0) = Θ1(u0) · Θ1(x) · 2 Θ1(v0) = 0, (because Θ1(x) = 0). By similar calculations, we can see that, ∀u, v ∈ U with every degree different from −1, 0, 1, Θ1(u · v) = 0. Now, we focus on U0, so pick an element p = p(c, h) and calculate the corresponding value of EXP1. Since Θ1(p(c, h)) = p(Θ1(c), Θ1(h)) = p(diag(3, 3), diag(1, −1)), we can deduce that ∀p ∈ U0, Θ1(p(c, h) = 0) if and only if p(3, 1) = 0 and p(3, −1) = 0. Note that the corresponding ideal is not prime. Anyway, Θ1(p(c, h)) is a diagonal matrix with its eigenvalues p(3, 1) and p(3, −1), and the p(3,1) p(3,−1) p(3,1)+p(3,−1) matrix EXP1(p) = diag(e , e ) with determinant equal to e . By what sketched above, Θ1 acts as zero on U±2,U±3,.... So, we restrict our attention to U−1,U0,U1. Let us pick up in U−1 ⊕U0 ⊕U1 an element γ = yp−1(c, h)+p0(c, h)+xp1(c, h) 8 SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2 where the polynomials p(c, h), p0(c, h), p1(c, h) belong to U0. We want to calculate the exponential value of γ, as follows:
EXP1(γ) = EXP1 ((yp−1(c, h)) + (p0(c, h)) + (xp0(c, h))) =
= exp (Θ1(yp−1(h, c)) + Θ1(p0(c, h)) + Θ1(xp1(c, h))) =
= exp (Θ1(y)Θ1(p−1(c, h)) + Θ1(p0(c, h))) + Θ1(x)Θ1(p1(c, h))) .
Since the value of Θ1 calculated on any element in U0 is represented by a diagonal matrix, so Θ1(yp−1(c, h)), Θ1(p0(c, h)), Θ1(xp1(c, h)) can be respectively represented by the diagonal matrices diag(a−1, b−1), diag(a0, b0), diag(a1, b1), where ai, bi ∈ C, with i = −1, 0, 1. So, we have 0 0 0 1 EXP (γ) = exp · diag(a , b ) + diag(a , b ) + · diag(a , b ) = 1 1 0 −1 −1 0 0 0 0 1 1 0 0 0 b1 = exp + diag(a0, b0) + = a−1 0 0 0 a b = exp 0 1 . a−1 b0
Thanks to these calculations, we can easily find the EXP1 of xp1(h, c): indeed, EXP1(xp1(h, c)) = 0 b 0 b 1 b exp(Θ (xp (h, c))) = exp 1 = I + 1 = 1 , (because the square 1 1 0 0 2 0 0 0 1 of the matrix x, so of xp1(c, h) is null). So, EXP1(xp1(h, c)) = I2 + Θ1(xp1(h, c)). A similar property holds for yp−1(c, h), in fact, EXP1(yp−1(c, h)) = exp(Θ1(yp−1(c, h))) = 0 0 1 0 exp = . a−1 0 a−1 1
7. Other exponential maps In this section, we define other exponential maps over U by using finite dimensional representations of sl2(C), that is, finite dimensional sl2(C)-modules ([6] 1.2). (All our modules will be left modules). First, recall that by Weyl’s theorem, any finite dimensional representation of sl2(C) can be decomposed as a direct sum of simple sl2(C)-modules ([6] 1.8.5). For every positive integer λ, there exists a unique (up to isomorphism) simple sl2(C)- module Vλ of dimension λ+1; Vλ can be described as the C-vectorspace of all homogeneous polynomials of degree λ with coefficients in C and variables X and Y (see [8, Chapter λ λ−1 λ−1 λ 5]). We decompose Vλ with respect to the basis of monomials X ,X Y,...,XY ,Y , λ λ−i i Vλ = ⊕i=0C[X Y ]. The representation fλ of sl2(C) can be described as follows: ∂ x acts as X ∂Y ∂ y acts as Y , ∂X ∂ ∂ h acts as X − Y . ∂X ∂Y So, for 0 < i ≤ λ, the basis element Xλ−iY i is shifted to the left, by the action of x, sent to i · X(λ−i)+1Y i−1 and for i = 0, Xλ is sent by x to 0. For 0 ≤ i < λ, the basis element Xλ−iY i is shifted to the right, by the action of y, sent to (λ−i)X(λ−i)−1Y i+1 and for i = λ, Y λ is sent by y to 0. Each subspace generated by Xλ−iY i is left invariant by the action EXPONENTIATIONS OVER THE UNIVERSAL ENVELOPING ALGEBRA OF sl2(C) 9 of h: Xλ−iY i is mapped to (λ − 2)Xλ−iY i (so the corresponding eigenvalue is equal to λ − 2.i). The C-vectorspace End(Vλ) coincides with the C-vectorspace Mλ+1(C) of all (λ + 1) × (λ + 1) matrices written with respect to a basis of eigenvectors for h. More precisely, through the representation fλ, the actions of x, y and h are represented respectively the following three (λ + 1) × (λ + 1) matrices Xλ+1, Yλ+1, Hλ+1, λ ∈ ω − {0}: 0 0 ... 0 0 1 0 ... 0 λ 0 ... 0 0 0 2 ... 0 0 λ − 1 0 (2) Xλ+1 = . . ,Yλ+1 = , . . λ . . 0 0 0 ... 0 . . 0 0 1 0
Hλ+1 = diag (λ, λ − 2,..., −λ + 2, −λ) .
Note that the operator norm of Xλ+1 (respectively Yλ+1) is equal to |λ|, as is the operator 2 norm of Hλ + 1. The norm of Θλ(c) is equal to |λ + 2λ|. On the other hand the F -norm of Xλ+1 is equal to λ(λ + 1)/2. For every positive integer λ, we have the following diagram.
Diagram 7.1. For any simple representation Vλ of sl2(C) of dimension λ + 1 (with λ ∈ ω − {0}), let us consider the representation map fλ : sl2(C)→Mλ+1(C), and the following (commutative) diagram determined by the universal property of U:
σ - sl2(C) U
f λ Θ λ ? © Mλ+1(C) where σ is the canonical mapping (which is a Lie-algebra homomorphism) from sl2(C) to U and Θλ is the (unique) algebra homomorphism from U to Mλ+1(C) sending 1 to 1 making the diagram commutes. Using the commutativity of the above diagram, we obtain that the images of x, y, h by the representation map Θλ : U → Mλ+1(C) coincide with their images by the representation map fλ, and so are equal to the matrices Xλ+1,Yλ+1,Hλ+1, (see (2)). The image by Θλ of the Casimir element c in U is given by the following calculation. 2 2 Θλ(c) = Θλ(2x · y + 2y · x + h ) = 2Θλ(x) · Θλ(y) + 2Θλ(y) · Θλ(x) + (Θλ(h)) = 2 = 2Xλ+1 · Yλ+1 + 2Yλ+1 · Xλ+1 + Hλ+1 = (3) = diag λ2 + 2λ, . . . , λ2 + 2λ .
By the technique used for defining the exponential map EXP1 from U to GL2(C), we can define the exponential map EXPλ for every positive integer λ, as follows. 10 SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2
Definition 7.1. Let λ ∈ ω − {0}. The exponential map EXPλ over U is obtained by composing Θλ with the natural exponential map exp from Mλ+1(C) to GLλ+1(C) (see section 5):
EXPλ(u) = exp (Θλ(u)) ∀u ∈ U.
Proposition 7.2. ∀λ ∈ N − {0}, the map EXPλ is surjective.
Proof. Since exp is surjective from Mλ+1(C) to GLλ+1(C), it suffices to prove that Θλ : U → Mλ+1(C) is surjective. The latter is deduced directly by Jacobson density theorem [14, Section 2.2]. For convenience of the reader, we indicate below the proof. Let Vλ be the irreducible representation of sl2(C) of dimension λ + 1. As representation ∼ of U, we know by Schur’s lemma, that EndU (Vλ) = C. Consider φ ∈ EndC(Vλ) (= Mλ+1(C)). Then by Jacobson density theorem we get that, for each finite subset of elements Vm v1, . . . , vλ+1 ∈ Vλ, that there exists u ∈ U such that i=1 ( φ(vi) = Θλ(u).vi) . We can easily calculate (as matrices in GLλ+1(C)) the values by EXPλ of x, y, h, c, using on one hand that Θλ(x), Θλ(y) are nilpotent matrices (in Mλ+1(C))), and on the other hand that Θλ(h), Θλ(c) are diagonal matrices. 2 λ Xλ+1 Xλ+1 EXPλ(x) = exp(Θλ(x)) = exp(Xλ+1) = Iλ+1 + Xλ+1 + 2 + ... + λ! ; 2 λ Yλ+1 Yλ+1 EXPλ(y) = exp(Θλ(y)) = exp(Yλ+1) = Iλ+1 + Yλ+1 + 2 + ... + λ! , λ λ−2 −λ+2 −λ EXPλ(h) = exp(Θλ(h)) = exp(Hλ+1) = diag(e , e , . . . , e , e ); 2 2 λ2+2λ λ2+2λ EXPλ(c) = exp(Θλ(c)) = exp(diag(λ + 2λ, . . . , λ + 2λ)) = diag(e , . . . , e ).
Furthermore, we easily see that EXPλ satisfies the properties properties of the matrix exponential exp described by Proposition 5.1.
Proposition 7.3. Let λ ∈ N − {0}. Then (U, EXPλ, GLλ+1(C)) is an exponential C- algebra. More precisely, we have the following properties. Let u, v ∈ U and let a, b ∈ C, then :
(i) EXPλ (0U ) = Iλ+1, where 0U denotes the identity element (with respect to the ad- dition) in U. (ii) EXPλ (a · u) · EXPλ (b · u) = EXPλ ((a + b) · u); (iii) EXPλ (u) · EXPλ (−u) = Iλ+1; (iv) for u and v commuting, EXPλ (u + v) = EXPλ (u) · exp(v); −1 −1 (v) for an invertible element v in U, EXPλ (vuv ) = Θλ(v)EXPλ (u)Θλ(v) ;
Proof. (i) By definition of EXPλ, EXPλ (0U ) = exp(Θλ(0U )) = exp(0λ) = Iλ+1 (see Proposition 5.1(i)). (ii) EXPλ (au) · EXPλ (bu) = exp(Θλ (au)) · exp(Θλ (bu)) = exp(a Θλ (u)) · exp(b Θλ (u)). Since Θλ(u) ∈ Mλ+1(C) and Proposition 5.1, (ii) can be applied, then exp(aΘλ (u)) · exp(bΘλ(u)) = exp((a + b)Θλ(u)) = exp(Θλ((a + b) u)) = EXPλ((a + b)u). (iii) This follows immediately from the corresponding property for the matrix exponential. (iv) First, note that if u and v commute in U, then Θλ(u) and Θλ(v) commute also (for Θλ is a homomorphism from U to Mλ+1(C) for every λ). Thus, by using Proposi- tion 5.1 (iv) and the fact that Θλ is a homomorphism, we have: EXPλ(u) · EXPλ(v) = exp(Θλ(u)) · exp(Θλ(v)) = exp(Θλ(u) + Θλ(v)) = exp(Θλ(u + v)) = EXPλ (u + v). (v) The map Θλ is a morphism of associative rings, so if an element v ∈ U is invertible, then so is Θλ(v).The result follows immediately by the corresponding property for the matrix exponential. EXPONENTIATIONS OVER THE UNIVERSAL ENVELOPING ALGEBRA OF sl2(C) 11
Note that since the Casimir element is central in U, its image Θλ(c) is central in Θλ(U) ⊆ Mλ+1(C), so for any u ∈ U, we get by Proposition 5.1 that exp(Θλ(c) + Θλ(u)) = exp(Θλ(c)) · exp(Θλ(u)). So, EXPλ(c + u) = EXPλ(c) · EXPλ(u). As a direct consequence of the definition of the map EXPλ, we observe that u ∈ Ker(EXPλ) if and only if Θλ(u) ∈ Ker(exp). So in order to describe Ker(EXPλ), we should say as much as possible about Θλ(u) for u ∈ U.
Proposition 7.4. Decompose U = ⊕m∈ZUm. The representation map Θλ sends: (i) an element u0 of U0 onto a diagonal matrix, (ii) an element um ∈ Um, m > 0, onto an upper triangular matrix if 0 < m ≤ λ, otherwise (when m ≥ λ + 1) Θλ(um) = 0λ+1. (iii) an element um ∈ Um, m < 0, is mapped to a lower triangular matrix, if −λ ≤ m ≤ −1 and, otherwise, for m ≤ −λ − 1, to the zero matrix 0λ+1.
Proof. (i) Let u0 ∈ U0 − {0}; so u0 is of the form p(c, h) with p(x1, x2) ∈ C[x1, x2], where x1 and x2 are two commuting variables. We know that Θλ(p(c, h)) = p(Θλ(c), Θλ(h)), where Θλ(c) and Θλ(h) are the diagonal matrices described respectively by 3 and 2. Since any algebraic operation on diagonal matrices concerns just their diagonal entries, then for any polynomial p(x1, x2) ∈ C[x1, x2], we have that: 2 2 (4) Θλ(p(c, h)) = diag p λ + 2λ, λ , . . . , p λ + 2λ, −λ (∈ Mλ+1(C))
m (ii) For the positive integer m, let um be an element in Um of the form um = x · u0 where the 0-component u0 = p(c, h) as above. On one hand, suppose that m ≤ λ. By using the fact that Θλ is a homomorphism and the values of Θλ(x) and Θλ(p(c, h)) (described m m by (2) and (4) respectively) we have that Θλ(um) = Θλ(x · u0) = Θλ(x) · Θλ(u0) = m 2 2 Xλ+1 · diag p λ + 2λ, λ , . . . , p λ + 2λ, −λ , so Θ(um) is represented by the strictly upper triangular matrix with ?l ∈ C, 1 ≤ l ≤ (λ + 1) − m. 0 0 ?1 0 ... 0 0 0 0 ?2 ... 0 . . . . (5) . . 0 ?m . . . . 0 0 0 0 ... 0 0
On the other hand, assume that m ≥ λ+1. Since Θλ(x) is a nilpotent matrix, we can easily m see: Θλ(um) = Θλ(x) · Θλ(u0) = 0.
(iii) Similarly, we can repeat the same argument for any element um, with m < 0, of the m form y · u0. So, for −λ ≤ m ≤ −1 the image by Θλ of um, is a lower triangular matrix of the form 0 0 0 0 ... 0 0 0 0 0 ... 0 ?1 0 0 0 ... 0 (6) 0 ?2 0 0 ... . . . . 0 0 ... 0 ?−m ... m If m ≤ −(λ + 1). we have Θλ(um) = Θλ(y · α0) = 0. 12 SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2
Remark 1. An element u0 ∈ U0 belongs to the kernel of EXPλ if and only if ^ 2 (7) p λ + 2λ, λ − 2j ∈ 2πiZ 0≤j≤λ
In fact, for u0 = p(c, h) the diagonal matrix Θλp(c, h) belongs to Ker(exp) if and only if their diagonal entries described by (4) belongs to Ker(e) = 2πiZ.
Proposition 7.5. EXPλ maps an element u of U into SLλ+1(C) whenever (8) tr(Θλ(u)) ∈ 2πiZ. In particular, if u ∈ ⊕m6=0Um, then its image by EXPλ lies always in SLλ+1(C). Proof. For the first statement, it is enough to apply property (vi) of Proposition 5.1, so for any u ∈ U, the determinant of exp(Θλ(u)) equals 1 if the trace of Θλ(u) belongs to Ker(e) = 2πiZ. As to the second claim, first we can note that the map EXPλ maps x, y and their powers into SLλ+1(C), because their images by Θλ are matrices of trace 0. We get the same results m m with x (respectively y ). Since the subalgebra U0 is sent to the subalgebra of diagonal m matrices in Mλ+1(C), the image of an element αm = x · α0 in Um by Θλ is a matrix of trace 0 (as illustrated by (5)) and so its matrix exponential has determinant 1. The same m argument holds where αm = y · α0 (for negative m). Since the sum of matrices of trace 0 has trace 0, an element of ⊕m6=0Um is sent by EXPλ to SLλ+1(C). When we restrict Proposition 7.5 to any element u0 of U0, where u0 = p(c, h) (for some polynomial p(x1, x2) ∈ C[x1, x2]) the condition (8) means that the sum of eigenvalues of P 2 Θλ(u), 0≤j≤λ p(λ + 2λ, λ − 2j), has to belong to 2πiZ. Pd l P Pbd/2c 2 2l Put p(x1, x2) = l=0 ql(x1)x2, then 0≤j≤λ l=0 ql(λ + 2λ)(λ − 2j) = Pbd/2c 2 P 2l l=0 qj(λ + 2λ)[ 0≤j≤λ(λ − 2j) ]. Now, let us assume that u is in the kernel of T EXP , for some λ . Then, 0 λ∈Z; λ>λ0 λ 0 2 p(λ + 2λ, λ − j) ∈ 2πiZ, for all |λ| > λ0 and 0 ≤ j ≤ λ. In the remainder of this section, we will give a partial answer to the question of which elements u of U are such that Θλ(u) ∈ suλ+1 ∗ Recall that suλ+1 := {A ∈ Mλ+1(C): A = −A, tr(A) = 0}, and SUλ+1 := {X ∈ ∗ ∗ GLλ+1(C): X · X = Iλ+1, det(X) = 1}, where X denotes the conjugate transpose of X; it is a compact Lie group. Coming back first to the case λ = 1, it is well-known that the exponential map exp (defined in M2(C)) restricted to sl2(C) does not map it surjectively to its Lie group SL2(C) ([21] page 38). However if we restrict to the R-subalgebra su2, exp is surjective onto the (compact) Lie group SU2(C) (see Lemma 2.a in section 2 of [21]). We have the following decomposition: SL2(C) = SU2(C).B, where B is the subgroup of triangular matrices with determinant 1 and positive real diagonal entries ([21] page 39). The surjectivity property of exp holds if one replaces su2 with suλ+1 and SU2 by SUλ+1 (see Corollary 2 in [21]). P 2 Let u ∈ U0, so u = p(c, h). So, Θλ(u) ∈ suλ+1, if j p(λ + λ, λ − 2j) = 0 and for all −λ ≤ j ≤ λ, p(λ2 + λ, λ − 2j) = −p¯(λ2 + λ, λ − 2j). The last condition occurs, for instance if p(x1, x2) is the multiple by the complex number i of a polynomial with real coefficients. P ` ` Now consider elements u ∈ ⊕m6=0Um, namely u = `>0(p`(c, h) · x + y · q`(c, h)) with 2 p`, q` ∈ C[h, c]. Then the condition under which Θλ(u) ∈ suλ+1 is that (λ − j) · q`(λ + EXPONENTIATIONS OVER THE UNIVERSAL ENVELOPING ALGEBRA OF sl2(C) 13
2 λ, λ − 2j) = (−λ + j) · p`(λ + λ, λ − 2j), for all −λ ≤ j ≤ λ. Given a polynomial p`, we can always find a polynomial q` (of degree ≤ λ − 1) meeting these λ conditions, using Lagrange interpolation theorem. 0 0 So, given u ∈ ⊕m>0Um, there exists u ∈ ⊕m<0Um such that Θλ(u + u ) ∈ suλ+1. 8. Exponentiations and ultraproducts
We will be considering a non principal ultraproduct of the Lie algebras Mλ+1(C), λ ∈ ω. Namely, let V be a non-principal ultrafilter on ω and consider the corresponding ultraprod- Q Q ucts Mλ+1(C) and GLλ+1(C). V V Q ByLos’s theorem, the structure ( V Mλ+1(C), +, −, 0, [·, ·]) is a Lie algebra over C or ∗ Q over C := V C, which is infinite-dimensional. We first observe the following.
Proposition 8.1. (i) If u0 is any element of U0 − {0}, then there exists λ0 such that for all λ ≥ λ0, we have Θλ(u0) 6= 0. (ii) For any u ∈ U − {0}, there exists λ0 such that for all λ ≥ λ0 we have Θλ(u) 6= 0. Proof. (i) Let u0 ∈ U0 − {0}; so u0 is of the form p(c, h) with p(x1, x2) ∈ C[x1, x2], where x1 and x2 are two commuting variables. The claim can be deduced directly from [11, Lemma 19]. For convenience of the reader, we repeat the argument here. We argue by contradiction. V 2 2 Assume that 0≤j≤λ p(λ +2λ, λ−2j) = 0. First, we choose λ such that p(λ +2λ, x2) 6= 0, so as a polynomial in j, p(λ2 + 2λ, λ − 2j) is non-trivial of degree k and so the number of roots is bounded by k. So, if we choose λ big enough, we will always find j such that 2 p(λ + 2λ, λ − 2j) 6= 0. Therefore, Θλ(p(c, h)) 6= 0 for some λ. th (ii) Let u ∈ U − {0}, then there exists m ∈ Z such that its m component um 6= 0. m Assume that m ≥ 0 and that m is minimal such. Let um = x u0, where u0 ∈ U0. Let Pd l p(x1, x2) ∈ C[x1, x2] be such that u0 = p(c, h). Write p(x1, x2) = l=0 qi(x1)x2. We can find (explicitly) an interval [−r; r] in R such that all the roots of the polynomial qd(x1) 0 0 2 are in that interval. Let r = max{r, d}. Then if λ > r , then qd(λ + 2λ) 6= 0 and so the Pd 2 l polynomial l=0 ql(λ + 2λ)x2 has less than d roots and among the λ + 1 elements of the form (λ−2j) where 0 ≤ j ≤ λ, we have such j with the property that p(λ2+2λ, (λ−2j)) 6= 0. Since the images of any homogeneous components Um with −λ ≤ m ≤ λ are in direct sum and Θλ(um) 6= 0, then we have Θλ(u) 6= 0. 2
Define the obvious Θ := [Θλ] from U to the ultraproduct of the Mλ+1(C), over any non-principal ultrafilter V on ω. By Proposition 8.1, the map Θ is an associative ring monomorphism. So, we get the following corollary. Corollary 8.2. For any non-principal ultrafilter V on ω, U embeds in the associative Lie Q algebra V Mλ+1(C). 2 Recall that U is a left and right Ore domain, so it has a left and right field of fractions which embeds in the ring U 0 of definable scalars of U. This ring U 0 has been shown to be von Neumann regular by I. Herzog ([11]), equivalently every left (right) principal ideal 0 is generated by an idempotent. Moreover, since any Vλ is also a U -module, we can send 0 Q r ∈ U in the direct product λ∈ω Mλ+1(C) by sending it in each factor to the element of Mλ+1(C), representing its action on each Vλ. 0 Then, [16] explicitly identifies certain idempotents of U of the form eu, u ∈ U, cor- responding to the projections on ker(Θλ(u)) on Vλ, λ ∈ ω. For instance ex is the pro- jection on the highest weight space of Vλ. When u ∈ U0, so of the form p(c, h), with 14 SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2 p(x1, x2) ∈ C[x1, x2], they call p standard if there are only finitely many λ such that p(λ2 + 2λ, λ − 2.j) = 0 for some 0 ≤ j ≤ λ (and non-standard otherwise). Note that if Q u = p(c, h) with p standard, then [Θλ(u)]V is invertible in V Mλ+1(C). (Note that the converse holds if [Θλ(u)]U is invertible with respect to any non-principal ultrafilter U). Let now u = p(c, h) ∈ U0 be such that p is non-standard, so for some non-principal Q ultrafilter V the action of eu in the ultraproduct V Vλ will be a non invertible element of the form [(diag(0,..., 1,..., 0, 1,..., 0)] 6= 0, where the number of possible 0’s is bounded by the degree of p with respect to the second variable.
Q Q We know that Θ is a surjection from V U to V Mλ+1(C) (see the proof of Proposition 7.2). Then, we will compose with the map Y Y Exp : Mλ+1(C) → GLλ+1(C) : [Aλ]V → [exp(Aλ)]V . V V ∗ Q Q So, by composing with [Θλ]V , we get a map EXP = Exp[Θλ]V from V U to V GLλ+1(C), which is surjective. The kernel of that map is in bijection with the kernel of Exp on Q V Mλ+1(C). Q Definition 8.1. Let EXP from U to V GLλ+1(C) be defined as follows: Y EXP : U → GLλ+1(C) : u → [EXP λ(u)]V . V Q Q Proposition 8.3. Both (U, EXP, GLλ+1(C)) and (U0, EXP, Diagλ+1(C)) are expo- V Q V nential C-algebras. Moreover we have that EXP (⊕m6=0Um) ⊂ SLλ+1(C), EXP (⊕m≥0Um) ⊂ Q Q V V UTλ+1(C), and EXP (U0) ⊂ V Diagλ+1(C). Proof: A direct application ofLosTheorem shows that EXP satisfies the properties stated for each EXPλ in Proposition 7.3. 2 Note that the above properties are independent of the non-principal ultrafilter V on ω. Question 8.1. What is the kernel of EXP?
It is the set of elements u such that for a subset of λ belonging to V, exp(Θλ(u)) = 1. So, the eigenvalues of Θλ(u) belong to 2πi·Z; does it translate into an independently interesting property of u ∈ U? For u0 ∈ U0, we have the following answer. Let p(x1, x2) ∈ C[x1, x2] 2 such that u0 = p(c, h). Then, for almost all λ and all 0 ≤ j ≤ λ, we have p(λ +λ, λ−2j) ∈ 2πi · Z.
Proposition 8.4. Let u := p(c, h) ∈ U0, with p(x1, x2) ∈ C[x1, x2]. Write p(x1, x2) in the form 2πi · q(x1, x2). Then, if u ∈ ker(EXP), then q(x1, x2) ∈ Q[x1, x2]. Pd k Proof: Let q(x1, x2) = k=0 qk(x1) · x2 and assume that q(c, h) ∈ ker(EXP ). Then, the V 2 set {λ ∈ ω : 0≤j≤λ q(λ + 2λ, λ − 2j) ∈ 2πi · Z} ∈ V (?). 2 Set ck := qk(λ + 2λ) and consider the following system of linear equations, with z` ∈ Z, 0 ≤ ` ≤ n: 2 d 1 y0 y0 ··· y0 c0 z0 2 d 1 y1 y1 ··· y1 c1 z1 . = . . . . . . . 2 d 1 yn yn ··· yn cn zn EXPONENTIATIONS OVER THE UNIVERSAL ENVELOPING ALGEBRA OF sl2(C) 15
2 d 1 y0 y0 ··· y0 2 d 1 y1 y1 ··· y1 When n = d, the determinant of the (square) matrix . . 2 d 1 yd yd ··· yd is equal to Q (y − y ). So it is a non zero integer whenever the y ’s are d 0≤n1
Remark 2. We have a partial converse to the above proposition. Namely, let q(x1, x2) = Pd k Pdk h k=0 qk(x1) · x2, where each qk(x1) ∈ Q[x1], so can be written as 1/nk · h=1 zh · x1 + q0,k, where nk ∈ N − {0}, zh ∈ Z and q0,k ∈ Q. If, we assume in addition that each q0,k ∈ Z, then for some ultrafilter V, 2πi · q(c, h) ∈ ker(EXP). Indeed, let n = lcm{nk : 0 ≤ k ≤ d}. Then we choose an ultrafilter V containing 2n · ω. 2 Pdk 2 h So, if λ = 2n · m, for some m ∈ ω, qk(λ + 2λ) = n/nk · h=1 zh · (2n · m + 2m) + q0,k, 2 V 2 then qk(λ + 2λ) ∈ Z and so {λ ∈ ω : 0≤j≤λ q(λ + 2λ, λ − 2j) ∈ 2πi · Z} ∈ V.
Corollary 8.5. Let u := p(c, h) ∈ U0, with p(x1, x2) ∈ C[x1, x2]. Write p(x1, x2) in the Pd k form 2πi · q(x1, x2). Write q(x1, x2) = k=0 qk(x1) · x2, with qk(x) ∈ Q[x1]. Then, u ∈ ker(EXP ) for all non-principal ultrafilters on ω, if and only if q(x1, x2) ∈ Q[x1, x2] and for each 0 ≤ k ≤ d, qk(0) ∈ Z. 2
Proposition 8.6. Let u := p(c, h) ∈ U0, with p(x1, x2) ∈ C[x1, x2]. Write p(x1, x2) in the Q form 2πi · q(x1, x2). Then, if EXP(u) ∈ SLλ+1(C), then q(x1, x2) ∈ Q[x1, x2].
Pd k Proof: Let q(x1, x2) = k=0 qk(x1) · x2 and assume that the set {λ ∈ ω : EXPλ(q(c, h)) ∈ Pbd/2c 2 P 2·` SLλ+1(C)} ∈ V. Equivalently, {λ ∈ ω :[ `=0 q`(λ +2λ)· 0≤j≤λ(λ−2j) ] ∈ 2πi·Z} ∈ V (?). Set ck := qk(x1) and consider the following system of linear equations, with z` ∈ Z, 0 ≤ ` ≤ n: 16 SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2
2 d 1 y0 y0 ··· y0 c0 z0 2 d 1 y1 y1 ··· y1 c1 z1 . = . . . . . . . 2 d 1 yn yn ··· yn cn zn 2 d 1 y0 y0 ··· y0 2 d 1 y1 y1 ··· y1 When n = d, the determinant of the (square) matrix . . 2 d 1 yd yd ··· yd is equal to Q (y − y ). So it is a non zero integer whenever the y ’s are d 0≤n1
9. Comparison with Serre’s definition of an exponential map. ˆ ∞ n Recall that the completion U of U ([22]) is defined as the infinite product Πn=0U , where U n denotes the component of degree n of U (generated by all products of length ≤ n of ˆ P∞ n generators x, y of U); an element f ∈ U can be represented as n=0 fn, where fn ∈ U n (see [22] Part 1, chapter 4, paragraph 6). (Note that U differs in general from Un.) Denote by M the ideal of U generated by x, y and let Mˆ be the ideal of Uˆ generated ˆ P f n by M. For f ∈ M, J.-P. Serre defines expS by the usual formula expS(f) := n n! . exp takes Mˆ to 1 + Mˆ (see [22] Part 1, chapter 4, paragraph 7). (Similarly, one can define logS ˆ ˆ P∞ n+1 xn from 1 + M to M by logS(1 + x) := n=1(−1) n , obtaining that expS ◦ logS = 1 = logS ◦ expS (see Theorem 7.2, Chapter 4, Part 1 in [22].) P∞ ˆ P∞ Let f := n=0 fn ∈ U and assume that n=0 Θλ(fn) belongs to Mλ+1(C). Then, define ˆ P∞ Θ(f) := [ n=0 Θλ(fn)]V . Since, if u ∈ U, there exists a bound on the number of non-zero components, this map is always well-defined on the elements of U.
Proposition 9.1. For any u ∈ M, Θ(ˆ expS(u)) = EXP (u). Pk j n Pk n Proof: Now, let u ∈ M with u = j=1 uj, where uj ∈ U , then u := ( j=1 uj) . So, for each m, the m-component of expS(u) is a finite sum. Therefore Θ(ˆ expS(u)) is well-defined ˆ P un and Θ(expS(u)) = [ n Θλ( n! )]V = EXP (u). 2
10. A ?-norm on the universal enveloping algebra of sl2(C). Now, we would like to put a natural topology on U in such a way that EXP is continuous. ∗ Q As in the previous section, we fix a non-principal ultrafilter V on ω; let C := V C be ∗ a non principal ultrapower of the field (C, +, ·, −, 0). We equip C with the ultrapower of the standard complex conjugation, and in addition consider the ultraproduct of the various Frobenius norms. This takes values in the corresponding ultrapower of the reals, and satisfies the obvious modification of the norm axioms. By functoriality this norm comes formally from the ultraproduct of the hermitian sesquilinear forms. Finally,by taking ultraproducts of normed algebras we get a natural notion of a ?-normed algebra, satisfying a natural version of the Cauchy-Schwartz inequality if the component algebras do. Since k · kλ+1 is a norm on each Mλ+1(C), by the usual properties of an Q ultraproduct, we get a natural ?-norm k · k on V Mλ+1(C) EXPONENTIATIONS OVER THE UNIVERSAL ENVELOPING ALGEBRA OF sl2(C) 17
This in turn, by 5.6 induces a star norm on U. In the next proposition, we will give an estimate of the norm of u ∈ U in terms of a polynomial in λ, with coefficients in R.
Lemma 10.1. For each u ∈ U, there exist non-zero polynomials q1(.), q2(.) with coefficients 2 in R such that for λ sufficiently big, we have q1(λ) ≤ kΘλ(u)kF ≤ q2(λ) and so q1([λ]V ) ≤ kuk ≤ q([λ]V ).
P Proof: Let us examine the norm of Θλ(u) for any element of U. Let u = m∈ um m Z (where um ∈ Um and m ∈ Z). Moreover, for each m ≥ 0, each um = x · pm(c, h), and m u−m = y · p−m(c, h), where pm(x1, x2), p−m(x1, x2) ∈ C[x1, x2]. Assume that for some P k ∈ N, we have u = −k≤m≤k um, then we estimate kΘλ(u)k as follows. Assume λ ≥ k, then
2 X 2 X 2 ||Θλ(u)||F = ||Θλ( um)||F = || Θλ(um)||F m∈Z m∈Z −1 k X X 2 = || Θλ(um) + Θλ(u0) + Θλ(um)||F m=−k m=1 −1 k X 2 2 X 2 = ||Θλ(um)||F + ||Θλ(u0)||F + ||Θλ(um)||F m=−k m=1 −1 k X |m| 2 2 X m 2 = ||Θλ(y p−m(c, h))||F + ||Θλ(p0(c, h))||F + ||Θλ(x pm(c, h))||F m=−k m=1 −1 X |m| 2 2 = ||Θλ(y) · p−m(Θλ(c), Θλ(h))||F + ||p0(Θλ(c), Θλ(h))||F m=−k k X m 2 + ||Θλ(x) · pm(Θλ(c), Θλ(h))||F . m=1
Pdm j Then, we make the following estimate. Write pm(x1, x2) = j=0 qj(x1).x2. Let fj(x1) = qj (x1) Pdm j and write the roots of fj(x1).x as α1(x1), ··· , αdm (x1). Note that these roots qdm (x1) j=0 2 Pdm−1 2 Pdm 2 are all in a ball of radius Mm(λ) := 1+ j=0 |fj(λ +2λ)|; let Rm(λ) := j=0 |qj(λ +2λ)|. Qdm 2 2 Then pm(x1, x2) = qdm (x1). j=1(x2 − αj(x1)). We have |pm(λ + 2.λ, λ − 2i)| = |qdm (λ + Q 2 2 2λ)|. j |((λ − 2i) − αj(λ + 2.λ)|. Since the number of roots of pm(λ + 2λ, x2) is at most d , there is at least one integer in the interval [−λ; λ] at distance bigger than b λ c m dm 2d of all of these roots. So, |q (λ2 + 2λ)|2.b λ c m ≤ P |p (λ2 + 2.λ, λ − 2i)|2 ≤ dm dm −λ≤i≤λ m 18 SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2
2 2dm+1 2 2dm+1 Rm(λ) .(2.λ + 1) ≤ Rm(λ) .(3λ ). So we get on one hand,
−1 2 X 2.|m| X 2 2 X 2 2 ||Θλ(u)||F ≤ λ · |p−m(λ + 2λ, λ − 2i)| + |p0(λ + 2λ, λ − 2i)| m=−k −λ≤i≤λ −λ≤i≤λ k X 2m X 2 2 + λ · |pm(λ + 2λ, λ − 2i)| m=1 −λ≤i≤λ −1 k X 2.|m| 2 2dm+1 2 2d0+1 X 2m 2 2dm+1 ≤ λ · Rm(λ) .(3λ ) + R0(λ) .(3λ ) + λ · Rm(λ) .(3λ ). m=−k m=1 and on the other hand,
−1 2 X 2.|m| X 2 2 X 2 2 ||Θλ(u)||F ≥ (λ − k) . |p−m(λ + 2λ, λ − 2i)| + |pd0 (λ + 2λ, λ − 2i)| + m=−k −λ≤i≤λ −λ≤i≤λ k X 2m X 2 2 (λ − k) . |pm(λ + 2λ, λ − 2i)| m=1 −λ≤i≤λ
−1 2dm 2d0 X 2.|m| 2 2 λ 2 2 λ ≥ (λ − k) .|qdm (λ + 2λ)| .b c + |qd0 (λ + 2λ)| .b c + dm d0 m=−k k X λ 2dm (λ − k)2m.|q (λ2 + 2λ)|2.b c . dm d m=1 m
We can give an estimate of the degrees of q1 and q2. Namely, the degree of q2 is equal to max−k≤m≤k{2.deg(Rm)+2|m|+2.dm+1} and the degree q1 is equal to max−k≤m≤k{4.deg(qdm )+
2|m| + 2.dm}. (Note that 2.deg(qdm ) ≤ deg(Rm).) 2 Q The ultraproduct of the norms induces a topology both on V Mλ+1(C) (under which + and . are continuous) and on the U. A basis of neighbourhoods O of 0 (in U) is given ∗,+ by O := {u ∈ U : kuk ≤ }, where ∈ R − {0}. When we just consider them as topological spaces, we will call them ?-normed spaces. Q Q Then, we will consider the following topological subspaces GLλ+1(C) (dense in Mλ+1(C)) Q Q V V and V SLλ+1(C) which is a closed subspace of V Mλ+1(C)).
Lemma 10.2. (See [19] Corollary 6.2.32.) Let A, B ∈ Mλ(C), then ||exp(A + B) − exp(A)||λ ≤ ||B||λexp(||B||λ)exp(||A||λ). So, the exponential map is continuous on Mλ(C) and Lipschitz continuous on each compact subset of Mλ(C). 2 Q Proposition 10.3. Consider the ?-normed spaces (U, k · k) and ( Mλ+1(C), k · kλ+1). Q V The map EXP : U → GLλ+1(C) is continuous and maps bounded sets to bounded V Q sets. The image EXP (U0) is an abelian subgroup of GLλ+1(C) and EXP (⊕m6=0Um) Q V is included in V SLλ+1(C). Proof: The continuity is clear fromLosand the preceding lemma. EXPONENTIATIONS OVER THE UNIVERSAL ENVELOPING ALGEBRA OF sl2(C) 19
Note that if the sequence Aλ+1 ∈ Mλ+1(C) is bounded, namely the sequence kAλ+1kλ+1 is bounded, then the corresponding sequence kexp(Aλ+1)kλ+1 is bounded. Indeed, by Ak P∞ λ+1 kAλ+1kλ+1 definition, exp(Aλ+1) = k=0 (k)! , so the norm kexp(Aλ+1)kλ+1 ≤ e . The last statement follows from Proposition 7.5. 2 Q Note that a priori, EXP(U) is not a subgroup of GLλ+1(C); we will denote by Q V < EXP(U) > the subgroup generated by EXP(U) in V GLλ+1(C). The Campbell-Baker- Hausdorff formula which expresses for two matrices A, B, exp(A) · exp(B) as exp(C) where C is expressed as an infinite series in commutators in A and B, can be translated back with u and v in place of A and B to express EXP(u) · EXP(v) in terms of an infinite series in u, v ([21] section 1.3). Does < EXP(U) > have finite width with respect to EXP(U), namely does there exist a finite number k such that every element of < EXP(U) > can be written as a product of k elements of EXP(U)? x ∗ We consider the field R := (R, +, ., 0, 1, e ), and we denote by R a non principal ultra- power of R with respect to the ultrafilter V on ω. We will extend the exponential map EXP ∗ ∗ to U ⊗ R as follows. Let u ∈ U and s := [rλ]V ∈ R with rλ ∈ R, then EXP (u ⊗ s) := P P Exp[rλ.θλ(u)]V = Exp[θλ(rλ.u)]V and EXP ( ui ⊗ si) := Exp[θλ( ui.ri,λ)]V , where P i i si := [ri,λ]V . Note that i ui.ri,λ ∈ U. This is well defined. We will say that a topological group G is ?-path connected if given any two elements ∗ ∗ h0, h1 ∈ G, there is a continuous map g from [0; 1] := R ∩ [0; 1] to G with g(0) = h0 and g(1) = h1. Q Proposition 10.4. The subgroups < EXP(U) > and EXP(U0) of V GLλ+1(C) (re- ∗ ∗ spectively < EXP(U ⊗ R ) > and EXP(U0 ⊗ R ) are topological groups. Moreover, < ∗ ∗ EXP(U ⊗ R ) > and EXP(U0 ⊗ R ) are ?-path connected. Q Proof: First note that V GLλ+1(C) is a topological group as an ultraproduct of topological ∗ ∗ groups. So, the subgroups < EXP(U) >, EXP(U0), < EXP(U ⊗ R ) > and EXP(U0 ⊗ R ) are topological subgroups. ∗ ∗ The groups < EXP(U ⊗ R ) > and EXP(U0 ⊗ R ) are ?-path connected. We only ∗ ∗ prove that < EXP(U ⊗ R ) > is ?-path connected. Let g0, g1 ∈< EXP(U ⊗ R ) >. Then we can write g1 = EXP(u1) · ... · EXP(un) and g0 = EXP(v1) · ... · EXP(vm), ∗ where u1, ··· , un, v1, ··· , vm ∈ U ⊗ R . So, g1 = g0 · EXP(y1) · ... · EXP(yk), for some ∗ ∗ y1, ··· , yk ∈ U ⊗ R . Let t ∈ [0; 1] and set g(t) = g0 · EXP(t · y1) · ... · EXP(t · yk), so ∗ g(0) = g0 and g(1) = g1. Let us denote the set {g ∈< EXP(U) > : ∃t ∈ [0; 1] g =
EXP (ty1) · ... · EXP(t · yk)} by Cg0,g1 . First, let us check that the map from [0; 1]∗ to EXP(U), sending t to EXP(tu) is contin- ∗ uous at t1 ∈ [0; 1] . ∗ Let ∈ [0; 1] , then we have to find η such that if |t0 − t1| < η, then kEXP(t0 · u) − EXP(t1 · u)k ≤ . We have EXP(t0 · u) − EXP(t1 · u) = EXP(t1 · u) · [EXP((t0 − t1) · u) − 1]. So, kEXP(t0 · u) − EXP(t1 · u)k ≤ kEXP(t1 · u)k · kEXP((t0 − t1) · u) − 1]k. Now, kEXP((t0 − k((t −t )·u)k t1) · u) − 1]k ≤ |(t0 − t1)| · kuk · e 0 1 . Then we use the fact that the product (possibly non commutative) of two continuous functions is continuous (∗). So, by induction on n, we may deduce that the map sending t to EXP(t · y1) · EXP(t · y2) · ... · EXP(t · yk) is also continuous. Now suppose < EXP(U) > is the disjoint union of two open sets U1 and U2. Denote the intersection of U1 (respectively U2) with Cg0,g1 by O1 (respectively O2). The inverse image ∗ of O1 and O2 gives rise to a partition of [0; 1] , which is a contradiction. 20 SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2
For convenience of the reader, let us prove (∗). Let f(t), g(t) be two continuous maps on the interval [0; 1]∗ and assume one of them is bounded. Then consider the map sending t to the product f(t) · g(t); let us show it is continuous at t1, assuming that f is bounded. Estimate the difference: f(t) · g(t) − f(t1) · g(t1) = (f(t) − f(t1)) · g(t1) + f(t) · (g(t) − g(t1)). So, kf(t) · g(t) − f(t1) · g(t1)k ≤ k(f(t) − f(t1))k · kg(t1)k + kf(t)k.k(g(t) − g(t1))k. Note that the map sending t to EXP (tu) is bounded. Indeed, kEXP (tu)k ≤ ekt·uk ≤ e|t|·kuk ≤ ekuk. 2
11. The asymptotic cone Q In the previous section, we embedded U in a ?-normed space, namely V Mλ+1(C). Here, we will embed U into a complete metric space (with an R-valued metric) which will be the asymptotic cone associated with the family of normed algebras Mλ+1(C), λ ∈ ω, and a non-principal ultrafilter V on ω. We will first endow each Mλ+1(C) with a new norm scaled down by λ; this norm differs from the norms we previously introduced in the fact that the norms of Θλ(x), Θλ(y), Θλ(c), Θλ(h) will be a multiple of λ (see Proposition 11.2). Even though they didn’t name it asymptotic cone, it was introduced by L. van den Dries and A. Wilkie when they revisited Gromov’s proof that a finitely generated group of poly- nomial growth is nilpotent-by-finite. Given a group of polynomial growth, M. Gromov associated a converging sequence of discrete metric spaces scaled down by a sequence of well-chosen natural numbers. Then, van den Dries and Wilkie associated with any finitely generated group G a limited ultraproduct of discrete metric spaces quotiented out by in- finitesimals. This space is usually denoted by Cone(X, V), where X is a metric space associated with G and V a non principal ultrafilter on ω, note that Cone(X, V) may de- pend on V (see for instance [15], [5]). The advantage of using an ultraproduct construction is that one can easily transfer certain properties from the factors.
First, we introduce the map φ from Mλ+1(C) to N, sending A ∈ Mλ+1(C) to the number of non-zero coefficients of A. Of course, φ(A) = 0 iff A = 0. Let us check that (1) φ(A + B) ≤ φ(A) + φ(B), (2) φ(A · B) ≤ φ(A) · φ(B)
We denote the ij coefficient of A + B by (A + B)ij. We have that if (A + B)ij 6= 0, then either Aij 6= 0 or Bij 6= 0. P Let C := A · B, then Cij = k Aik · Bkj and so Cij 6= 0 implies that for some k, Aik 6= 0 and Bkj 6= 0. We prove the second claim by induction on the number φ(C). For φ(C) = 1, it is clear. By induction suppose that for any 1 ≤ n ≤ m, if φ(C) = n, then for some 2-tuple (k1, k2) with k1 ≥ 1, k2 ≥ 1, such that φ(A) ≥ k1 and φ(B) ≥ k2 and n ≤ k1 · k2. Assume now that φ(C) = m + 1, so there are m + 1 tuples (i, j) with Cij 6= 0. For each of these tuples, there are two tuples (i, k), (k, j) such that Aik 6= 0 and Bkj 6= 0. By induction corresponding to the first m non-zero tuples, we know that there are k1 (respectively k2) non-zero coefficients of the matrix A (respectively of the matrix B) which are non-zero and such that m ≤ k1.k2. Corresponding to the m + 1 non-zero coefficient of C, there exists another non-zero coefficient of either A or B and so either φ(A) ≥ k1 + 1, or φ(B) ≥ k2 + 1, so m + 1 ≤ min{(k1 + 1) · k2, k1 · (k2 + 1)}. So, this map φ defines a norm on Mλ+1(C), that we will denote by k · kc,λ+1. Q k·kc,λ+1 Q∗ k·kc,λ+1 In the ultraproduct V (Mλ+1(C), λ ), we consider the set V (Mλ+1(C), λ ) of elements [aλ] such that for some natural number N, we have {λ ∈ ω : kaλkc,λ ≤ N ·λ} ∈ V. Then we quotient out this set by the equivalence relation ∼ defined by [aλ]V ∼ [bλ]V if EXPONENTIATIONS OVER THE UNIVERSAL ENVELOPING ALGEBRA OF sl2(C) 21
kaλ−bλkc,λ ( ) →V 0. Let us denote the equivalence class of an element by [aλ]∼ and by st λ Q the standard part of an element of V R whose absolute value is bounded by some natural number. Q∗ k·kc,λ+1 On XV := V (Mλ+1(C), λ )/ ∼, we define the following distance with values in ≥0 R . kaλ−bλkc,λ Let a := [aλ]∼ and b := [bλ]∼, then d(a, b) := st([ λ ]).
Lemma 11.1. The space (XV (C), d) is an infinite-dimensional complete metric space. Proof: The only thing which remains to be checked is the completeness of the space, but this follows from the countable saturation of the ultraproduct. 2 k·kc,λ+1 We will say that (XV (C), d) is the asymptotic cone associated with {(Mλ+1(C), λ ); λ ∈ N} and V.
Proposition 11.2. The universal enveloping Lie algebra U of sl2(C) embeds in (XV (C), d) via its embedding in the ultraproduct of the matrix rings. Proof: We proceed in two steps. Q∗ k·kc,λ+1 Firstly, we show that for any u ∈ U, [Θλ(u)] belongs to V (Mλ+1(C), λ ). This is direct by inspection of the proof of 5.4. Secondly, let u ∈ ⊕|j|≤mUj, then there exist d0, d1, . . . , dm, d−1, . . . , d−m such that for all Pm Pm P−m λ ∈ N, φ(Θλ(u)) = (λ−d0)+ j=1((λ−i)−di) = λ.m−(m(m+1))/2− i=1 di− i=−1 d−i. Again, this is seen by inspection of the proof of 5.4. So if [Θλ(u)] ∼ 0, then [Θλ(u)] = 0. 2
We will denote the image of U in (XV (C), d) by U∼.
Definition 11.1. A matrix (aij) in Mλ+1(C) is called a m-band matrix if there exists m such that for any 1 ≤ i, j ≤ λ + 1, we have aij 6= 0 implies that |i − j| ≤ m. (Namely, the non-zero entries of a m-band matrix are confined to a diagonal band comprising the main diagonal and the adjacent m diagonals on either side. The band-width is equal to 2.m + 1.
Proposition 11.3. Every element of U acts by left multiplication on the image U∼ of U in Q (XV (C), d), in a continuous way. More generally, any element [aλ]V ∈ V Mλ+1(C) acts by right multiplication on (XV (C), d), whenever there exists m independently of λ such that aλ is a m-band matrix.
Proof: Let u, v ∈ U. Let us show that [Θλ(u)].[Θλ(v)]∼ is well-defined. Namely, if [λ] ∈ Q V (Mλ+1(C) with [λ] ∼ 0, then Θλ(u) · λ ∼ 0. Assume that u ∈ ⊕|j|≤mUj, so Θλ(u) is a band matrix of width ≤ m. Namely Θλ(u)ij = 0 unless |i − j| ≤ m. So, if we denote by c Pλ+1 the matrix in Mλ+1(C) which is the product Θλ(u) · λ, then cij = k=1 Θλ(u)ik · kj. If we fix the matrix element kj, then there are at most 2m indices i such that cij 6= 0. Now since φ(λ) φ(c) φ(λ) [λ] ∼ 0, lim λ = 0. We have that φ(c) ≤ φ(λ) · 2m, so lim λ ≤ lim λ · 2m = 0. .λ This action is continuous. Let > 0, choose η := . Then for v1, v2 ∈ U∼, if φ(Θλ(u)) d(v1, v2) ≤ η, then d(u.v1, u.v2) ≤ . Indeed, we have kΘλ(u).Θλ(v1) − Θλ(u).Θλ(v2)kc,λ = φ(Θλ(u).Θλ(v1) − Θλ(u).Θλ(v2)) = φ(Θλ(u).(Θλ(v1) − Θλ(v2))) ≤ φ(Θλ(u)).φ(Θλ(v1) − Θλ(v2)). 2 Acknowledgments We thank Carlo Toffalori for his careful reading of the paper and his suggestions. 22 SONIA L’INNOCENTE1, ANGUS MACINTYRE, AND FRANC¸OISE POINT2
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L’Innocente Sonia, Department of Mathematics and Computer Science, University of Camerino, Via Madonna delle Carceri 9, 62032 Camerino (MC) Italy E-mail address: [email protected]
Angus MacIntyre, School of Mathematics, Queen Mary University of London, Mile End Road London E1 4NS, England E-mail address: [email protected]
Franc¸oise Point, Institut de mathematique,´ Universite´ de Mons, Le Pentagone, 20, place du Parc, B-7000 Mons, Belgium. E-mail address: [email protected]